Pressure trace of a tank fed by a compressor

[Sweepy]

G'Day All,

This is my first post so please let me know if I have completed this form incorrectly, or missed a point of etiquette etc...

1. Homework Statement
The problem is to determine the pressurisation rate of a tank being filled by a pipe connected to a compressor.

Assumptions:

• Pipe friction is negligible
• Pipe volumetric flow rate is known
• Tank volume is constant
• Air is the fluid moving through the system and is treated as an ideal gas
• The temperature of the fluid is constant

Homework Equations

Equation of State:
$$pV=nRT$$

Linking Relationship:
$$\dot{n}_{tank} = \dot{n}_{pipe}$$

The rate of molecular flow into the tank is that coming from the pipe

The Attempt at a Solution

Using the equation of state for both the tank and the pipe and taking the time derivative we have the following.

Tank:
$$\frac{d(pV)}{dt}=\frac{d(nRT)}{dt}$$
$$\dot{p}V=\dot{n}RT$$ as V and T are constant

Pipe:
$$\frac{d(pV)}{dt}=\frac{d(nRT)}{dt}$$
$$\dot{n} = \frac{p\dot{V}}{RT}$$
NB: I've assumed that the pressure is time-invariant which I am not sure is valid. Will return to this

Therefore, combining the two equations yields the following relationship for the system.

$$\dot{p}_{tank} = \frac{p_{pipe}\dot{V}_{pipe}RT_{tank}}{V_{tank}RT_{pipe}}$$
$$\dot{p}_{tank} = \frac{p_{pipe}\dot{V}_{pipe}}{V_{tank}}$$

Here is where I get stuck. If the volumetric flow rate ($\dot{V}$) is inversely proportional to the pressure within the tank (i.e. $\dot{V}\propto\frac{1}{p_{tank}}$; this is what I believe based on compressor performance curves), then the relationship can be re-written as

$$\dot{p}_{tank} = \frac{p_{pipe}}{V_{tank}}\frac{C}{p_{tank}}$$
Where C is a constant

So,
$$\dot{p}_{tank} = \frac{\alpha}{p_{tank}}$$
Where $\alpha=\frac{p_{pipe}C}{V_{tank}}$

This is an seperable 1st order ODE which we can re-write as

$$\frac{dp}{dt} = \frac{\alpha}{p}$$
$$p\cdot dp = \alpha\cdot dt$$
$$\int p\cdot dp = \int \alpha\cdot dt$$
$$\frac{p^{2}}{2} = \alpha t + C$$, where C is a constant
$$p = \pm \sqrt{2\alpha t + 2C}$$, we reject the '-' value as negative pressure is unphysical

Finally, we have

$$p = \sqrt{2\alpha t + 2C}$$

This plots a pressure-time trace that I would expect, that is, one which flattens over time as the pressure in the tank increases creating a resistance (back pressure) to the compressor. But this is based on the assumption of the volumetric flow rate being inversely proportional to the pressure in the tank which I do not know for sure.

Additionally, I am not sure about my earlier assumption that the pressure exiting the pipe is constant. My gut tells me that this should be equal to the pressure of the tank (excluding any wacky compressibility effects) as they are connected directly

If that were the case, and the pipe pressure was equivalent to the tank pressure, then the inverse proportionality of the volumetric flow rate would cancel with this new pipe pressure (both being equal to $p_{tank}$) and would result in a linear pressure trace.

This doesn't feel right to me intuitively, but I thought I would put it up here to get some outside comment on my thinking/methodology

Thanks in advance!

 

[BvU]

Hello Sweepy,

I see some internal inconsistencies popping up:

Using the equation of state for both the tank and the pipe

for the pipe the volumetric flow rate is known. Is it constant ?

I've assumed that the pressure is time-invariant which I am not sure is valid.

What pressure ? Surely not the pressure for which you want to find the time dependence ?

For the pipe:
If you have $pV = nRT$ and $\dot V$ is constant and given, $T$ is constant, you have $\dot n = {p \dot V\over RT}\ \$.

For the tank:
With the tank volume constant I would expect ${\displaystyle \dot p = {\dot n RT \over V }= {p\dot V\over V}}\ \$.

If you have no pump curve, that's the best you can do over a limited period of time.